Can we consider compact sets of Riemannian manifolds as ones of closed Rimanninan manifolds?

Question

Let $(M,g)$ be a $C^\infty$ Riemannian manifold of $n$ dimensional and suppose $\emptyset\neq K \stackrel{\mathrm{compact}}{\subset} M$. Then are there any neiborhood $\Omega\stackrel{\mathrm{open}}\subset M$ of $K$, closed $C^\infty$ Riemannian manifold $(N,h)$ of $n$ dimensional and isometric embedding $f:\Omega\to N$ ?

Answer 1

Yes. Here's a sketch of a proof. (The theorem references are from my Introduction to Smooth Manifolds, 2nd ed.)

First, let $f\colon M\to \mathbb R$ be a smooth exhaustion function, i.e., a function whose sublevel sets $M_c = f^{-1}((-\infty,c])$ are compact for all $c\in\mathbb R$. (Such a function exists by Prop. 2.28.) Because the sets $U_c = f^{-1}((-\infty,c))$ form an increasing open cover of $K$, there is some $C$ such that $K\subseteq U_C$.

By Sard's theorem (Thm. 6.10), there is some $c>C$ that is a regular value of $f$, and then $M_c$ is a regular domain in $M$ whose interior contains $K$ (Prop. 5.47). Since a regular domain is, in particular, a smooth manifold with boundary, we can form its double $D(M_c)$, which is a compact manifold without boundary containing a diffeomorphic copy of $M_c$ (Example 9.32). Using a partition of unity, it is straightforward to extend the given Riemannian metric on $M_c$ to a smooth Riemannian metric on all of $D(M_c)$.